What is the limit as x approaches 0 of (1+2x)^cscx(1+2x)cscx?

1 Answer
Dec 23, 2014

The answer is e^2e2.

The reasoning is not that simple. Firstly, you must use trick: a = e^ln(a).

Therefore, (1+2x)^(1/sinx) = e^u(1+2x)1sinx=eu, where
u=ln((1+2x)^(1/sinx)) = ln(1+2x)/sinxu=ln((1+2x)1sinx)=ln(1+2x)sinx

Therefore, as e^xex is continuous function, we may move limit:
lim_(x->0) e^u = e^ (lim_(x->0)u)

Let us calculate limit of u as x approaches 0. Without any theorem, calculations would be hard. Therefore, we use de l'Hospital theorem as the limit is of type 0/0.
lim_(x->0) f(x)/g(x) = lim_(x->0)((f'(x))/(g'(x)))
Therefore,
lim_(x->0) ln(1+2x)/sinx = 2/(2x+1)/cos(x) = 2/((2x+1)cosx) = 2

And then, if we return to the original limit e^ (lim_(x->0)u) and insert 2, we get the result of e^2,